Concrete mathematics : a foundation for computer science

4.9 PHI AND MU 139
= x f(x/m) x p(d) = x f(x/m)[m=l] = f(x).
m>l d\m lll>l
The proof in the other direction is essentially the same.
So now we can solve the recurrence (4.60) for a(x):
D,(x) = ; x Ad) lx/d.lll + x/d1
d>l
This is always a finite sum. For example,
Q(12) = ;(12.13-6.7-4.5+0-2.3+2.3
-1~2+0+0+1~2-1~2+0)
ZI 78-21-10-3+3-1+1-l = 46.
(4.62)
In Chapter 9 we’ll see how to use (4.62) to get a good approximation to m(x);
in fact, we’ll prove that
Q(x) = -$x2 + O(xlogx).
Therefore the function O(x) grows “smoothly”; it averages out the erratic
behavior of cp(k).
In keeping with the tradition established last chapter, let’s conclude this
chapter with a problem that illustrates much of what we’ve just seen and that
also points ahead to the next chapter. Suppose we have beads of n different
colors; our goal is to count how many different ways there are to string them
into circular necklaces of length m. We can try to “name and conquer” this
problem by calling the number of possible necklaces N (m, n).
For example, with two colors of beads R and B, we can make necklaces
of length 4 in N (4,2) = 6 different ways:
f-R\ /R\ fR\ c-R\ /R-\ c-B>
RR RR RB BB BB BB